áp dụng BĐT\(\frac{1}{x}+\frac{1}{y}>=\frac{4}{x+y}\)(x,y>0)
=>A=\(\frac{1}{xy}+\frac{2}{x^2+y^2}=\frac{2}{2xy}+\frac{2}{x^2+y^2}=2\left(\frac{1}{2xy}+\frac{1}{x^2+y^2}\right)>=\frac{2.4}{2xy+X^2+Y^2}=\frac{8}{\left(x+y\right)^2}=8\)
dấu bằng xảy ra khi x=y=1/2